Consider the window (of area 0.68 m2) of an airplane. If the pressure inside the plane is atmospheric pressure, while the pressure outside is 31 % of Patm, find the force on the window. Ignore the velocity of the airplane (i.e., ignore Bernoulli’s effect).
(1 -0.31)*Patm*(window area)= ___ N/m^2
Patm refers to the normal seal level pressure, 1.013*10^5 N/m^2
To find the force on the window, we need to calculate the pressure difference between the inside and outside of the airplane.
First, let's determine the pressure difference. The pressure outside the plane is given as 31% of atmospheric pressure (Patm). So, the pressure outside (Poutside) can be calculated as:
Poutside = 0.31 * Patm
The pressure inside the plane is the atmospheric pressure (Patm).
The pressure difference (ΔP) between inside and outside the plane is given by:
ΔP = Patm - Poutside
= Patm - 0.31 * Patm
= 0.69 * Patm
Now, we can calculate the force on the window using the pressure difference and the area of the window.
Force = Pressure difference * Area = ΔP * A
Given that the area of the window is 0.68 m^2, we can substitute the values into the equation:
Force = 0.69 * Patm * 0.68 m^2
So, to get the force on the window, we need to know the value of atmospheric pressure (Patm).